Deriving Biot-Savart's Law, kinda a math problem not really physics

  • Thread starter flyingpig
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  • #26
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But your theta is wrong, at least in my picture it is wrong.
 
  • #27
jhae2.718
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x is the green line, right? Then if theta is between r and y (the line ds is along), then x = r*sin(theta)
 
  • #28
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yeah, you had [tex]<sin\theta, cos\theta,0>[/tex]
 
  • #29
jhae2.718
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Your vector is in <x,y,z> form, right? Then x = rsin(theta), y = rcos(theta), and z = 0.
 
  • #30
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Wait you may be onto something...*goes and fix tex*
 
  • #31
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Okay! Tex is fixed on first page, made some other minor substition fixed. BUt my x^2 still remains..
 
  • #32
jhae2.718
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Use the approach you did in post #3...

I also TeX-ed up my solution; I'll post it in a little bit...
 
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  • #33
SammyS
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Okay! Tex is fixed on first page, made some other minor substitution fixed. But my x^2 still remains..
The last I looked, you did have inconsistencies between the way θ is drawn in your figure and the way it's used in some of your equations.


[tex]d\vec{s} = <0,d\vec{y},0>[/tex]  should be  [tex]d\vec{s} = <0,dy,0>[/tex]

[tex]\hat{r} = <\cos\theta,\sin\theta,0>[/tex]  should be  [tex]\hat{r} = <\sin\theta,\cos\theta,0>[/tex]

In general  [tex]\left|\vec{A}\times\vec{C}\right|=\left|\vec{A}\right|\left|\vec{C}\right|\sin\theta[/tex] so your cross product should have sin θ, not cos θ.

Jumping down to one of your equations for B:

[tex]\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-cos\theta d\vec{y} \hat{k}}{x^2 + y^2}[/tex]

After the above corrections and sin θ = x/r, you should have:

[tex]\vec{B} = \frac{\mu_0 I}{4\pi} \int \frac{-\sin\theta dy \hat{k}}{x^2 + y^2}[/tex]
[tex]=-\,\frac{x\mu_0 I}{4\pi}\,\hat{k}\, \int \frac{ dy }{(x^2 + y^2)^{3/2}}[/tex]

x comes out of the integral because you're integrating over y, not x.​
 
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  • #35
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This is going to sound unrelated, but how did you make that pdf that is so compatiable with LaTeX?
 
  • #36
SammyS
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Here's my solution:
View attachment 33843
I didn't do the integration, but my solution agrees with this.

But, I do think there is a problem with both solutions.

It's late, so I'll look at it tomorrow.

Below added in Edit:

The concern I have with my solution, is that the variable, y, was used in two distinct ways. The primary way was as the variable of integration. The other way y was used was as a coordinate for the point, P.

I intend to redo my solution using (x0, y0) for the coordinates of P.


.
 
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  • #37
jhae2.718
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This is going to sound unrelated, but how did you make that pdf that is so compatiable with LaTeX?

I used LaTeX :)

MikTeX 2.9 on Windows.

Source:
Code:
\documentclass[10pt,letterpaper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{fullpage}
\usepackage{graphicx}
\usepackage{float}
\usepackage{xfrac}
\newcommand{\custpic}[4]{\begin{figure}[H]
	\begin{center}
		\includegraphics[width=#2\textwidth]{#1}
		\vspace{-15pt}
		\caption{#3}
		\vspace{-20pt}
		\label{#4}
	\end{center}
\end{figure}}

\begin{document}
\custpic{pf5}{.75}{Problem diagram}{pf}

Start with Eq. (\ref{bs}), the Biot-Savart Law:
\begin{equation}\label{bs}
	d\vec{B} = \frac{\mu_0}{4\pi}\frac{id\vec{s} \times \hat{r}}{r^2}
\end{equation} 
From the diagram:
\begin{align*}
	r = \sqrt{x^2 + y^2} \\
	x = r\sin(\theta) \\
	y = r\cos(\theta) \\
	d\vec{s} = d\vec{y} 
\end{align*}
Substitute into Eq. (\ref{bs}):
\begin{equation}
	d\vec{B} = \frac{\mu_0}{4\pi}\frac{id\vec{y} \times \hat{r}}{x^2 + y^2}
\end{equation}
Taking the cross product, $id\vec{y} \times \hat{r} = dy\sin(\theta) (-\hat{k})$. Substitute in:
\begin{equation}
	d\vec{B} = \frac{\mu_0 i}{4\pi}\frac{dy\sin(\theta)}{x^2 + y^2} (-\hat{k})
\end{equation} 
From above, $\sin(\theta) = \dfrac{x}{r}$:
\begin{equation}
	d\vec{B} = \frac{\mu_0 i}{4\pi}\frac{xdy}{\left(x^2 + y^2\right)^{\sfrac{3}{2}}} (-\hat{k})
\end{equation} 
Find $\vec{B}$ by taking the integral over the infinite line:
\begin{equation}
	\vec{B} = \frac{\mu_0 i}{4\pi}\int\limits_{-\infty}^{\infty}\frac{xdy}{\left(x^2 + y^2\right)^{\sfrac{3}{2}}} (-\hat{k})
\end{equation} 
\begin{equation}
	\vec{B} = 2 \frac{\mu_0 i}{4\pi}\int\limits_{0}^{\infty}\frac{xdy}{\left(x^2 + y^2\right)^{\sfrac{3}{2}}} (-\hat{k})
\end{equation} 
Integrate (I used Wolfram$|$Alpha) to get:
\begin{equation}
	\vec{B} = 2 \frac{\mu_0 i}{4\pi}\left[\frac{y}{x\sqrt{x^2 + y^2}}\right]_{0}^{\infty} (-\hat{k})
\end{equation} 
The indefinite integral evaluates to:
\begin{equation}
	\vec{B} = \frac{\mu_0 i}{2\pi x} (-\hat{k})
\end{equation}
We can confirm this using Ampere's Law (left as an exercise to the reader...)
\end{document}
 
  • #38
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It's getting late here, I will recheck my math tomorrow, thank you to both of you very much. That sentence did not make any sense, but hopefully you understood my appreciation.
 
  • #39
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How did you save the file so perfectly? Whenever I saved it as a pdf, and I try reopen it it says it can't read it.
 
  • #40
SammyS
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...
The concern I have with my solution, is that the variable, y, was used in two distinct ways. The primary way was as the variable of integration. The other way y was used was as a coordinate for the point, P.
...
Simply use x0 for x, and use (y0 - y) for y.

[tex]\vec{B}(x_0,\,y_0)=-\,\frac{x_0\mu_0 I}{4\pi}\,\hat{k}\, \int \frac{ dy }{(x_0^2 + (y_0-y)^2)^{3/2}}[/tex]

I think you may find it handy to evaluate the integral from y0 to y .
 
  • #41
jhae2.718
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How did you save the file so perfectly? Whenever I saved it as a pdf, and I try reopen it it says it can't read it.

You need to run latex on it.

Simply use x0 for x, and use (y0 - y) for y.

[tex]\vec{B}(x_0,\,y_0)=-\,\frac{x_0\mu_0 I}{4\pi}\,\hat{k}\, \int \frac{ dy }{(x_0^2 + (y_0-y)^2)^{3/2}}[/tex]

I think you may find it handy to evaluate the integral from y0 to y .

That is an excellent point, and is something I just glossed over.
 
  • #42
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WHat dose that mean "run latex" over it...?
 
  • #43
jhae2.718
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If you have a version of LaTeX installed, open up the command prompt, change directory to the location where you have saved the file, and type:
Code:
pdflatex filename.tex
replacing filename with the actual filename.
 
  • #44
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I am guessing LaTeX is not the same thing as MikTeX lol
 
  • #45
jhae2.718
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MikTeX is a LaTeX distribution. http://docs.miktex.org/faq/faq.html

Try a LaTeX editor like TeXniCenter, Texmaker, etc to actually write/compile the code if you don't like the CLI.
 

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